Note that a freely falling object has a constant acceleration which is equal to gravity. Its sign varies depending on the object's direction. When going up, acceleration is positive and negative if the object is moving downward.

Also when thrown upwards, the object's velocity decreases. When veocity becomes zero, the object reaches its maximum height.

(a) To solve for the maximum height, let's use the formula:

`v_2^2=v_1^2+2as`

where v1 - initial velocity

v2 - final velocity

a - acceleration

s - change in height

Note that s is positive when it is in the same direction with the initial velocity.

Substitute `v_1=20 m//s` , `v_2=0` , and `a=-9.81m//s^2` .

`0=20^2-2(9.81)s`

`0=400-19.62s `

`19.62s=400`

`s=20.39`

Since the ball is 1.5m above the ground when released, the total height is:

`h=s+1.5 =20.39+1.5=21.89`

Hence, the maximum height of the ball is 21.89m.

(b) Since the velocity of the object becomes zero when it reaches its peak, then from its maximum height, the object starts to fall until it reaches the ground.

So to solve the velocity of the ball before it reaches the ground, use the same formula:

`v_2^2=v_1^2+2as`

Here, our initial velocity is `v_1=0`. The change in height is `s=21.89m` . And acceleration is `a=9.81m//s^2` .

`v_2^2=0+2(9.81)(21.89)`

`v_2^2=429.4818`

`v_2=20.72`

Hence, the velocity of the ball before it hits the ground is 20.72 m/s.